Wastes with Fuel Value

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Wastes with Fuel Value within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what wastes with fuel value describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Wastes with Fuel Value is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: wastes with fuel value.
Capstone Studio instructional photograph

Environmental Engineering — Wastes with Fuel Value: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Industrial Wastes
Average Heating Value (as fired)
Btu/lb kJ/kg
Waste gases:
Coke-oven 19,700 45,900
Blast-furnace 1,139 2,654
Carbon monoxide 579 1,349
Liquids:
Refinery 21,800 50,794
Industrial sludge 3,700–4,200 8,621–9,786
Black liquor 4,400 10,252
Sulfite liquor 4,200 9,786
Dirty solvents 10,000–16,000 23,300–37,280
Spent lubricants 10,000–14,000 23,300–32,620
Paints and resins 6,000–10,000 13,980–23,300
Oily waste and residue 18,000 41,940
Solids:
Bagasse 3,600–6,500 8,388–15,145
Bark 4,500–5,200 10,485–12,116
General wood waste 4,500–6,500 10,485–15,145
Sawdust and shavings 4,500–7,500 10,485–17,475
Coffee grounds 4,900–6,500 11,417–15,145
Nut hulls 7,700 17,941
Rice hulls 5,200–6,500 12,116–15,145

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

BOD exerted after t days — Wastes with Fuel Value

A wastewater has ultimate BOD L₀ = 283 mg/L and k = 0.19 /day (base e). How much BOD is exerted in 4 days?

Given

L₀ = 283 mg/L
k = 0.19 /day
t = 4 days

Find

BOD_t

Start with the thinking

BOD exertion is first-order and approaches L₀ asymptotically.
Check whether k is base e or base 10.

Step-by-step solution

First-order
$B O D_{t} = L_{0} (1 - e^{(} - k t))$
Exponent
$k t = 0.19 (4) = 0.76$
Exponential
$e^{(} - 0.76) = 0.4677$
Substituting
$B O D_{t} = 283 (1 - 0.4677) = 150.7 m g / L$
Remaining
$L_{0} - B O D_{t} = 132.3 m g / L$

Answer: BOD_4 ≈ 150.7 mg/L

Why the other options are there

132.3 mg/L (remaining reported as exerted)
215.1 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 2

Activated sludge F/M ratio, aeration time and sludge production — Wastes with Fuel Value

An activated sludge plant treats 16,806 m³/d with an influent BOD of 225 mg/L, effluent BOD 17 mg/L, MLSS 2,480 mg/L, and an aeration basin volume of 4,016 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 16 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 16,806 m³/d
S₀ = 225 mg/L, S = 17 mg/L
X = 2,480 mg/L, V = 4,016 m³
Y = 0.45, k_d = 0.06 d⁻¹, SRT = 16 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 16806 (225) / (4016 \times 2480) = 0.380 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 4016/16806 = 0.2390 d = 5.74 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.45 (16806) (225 - 17) / (1 + 0.06 \times 16) m g / d$
Evaluate
$P_{x} = 802.6 k g V S S / d a y$

Answer: F/M = 0.38 d⁻¹, τ = 5.7 h, sludge = 802.6 kg/d

Why the other options are there

F/M = 1,525 (basin volume omitted)
P_x = 1,573 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 3

Landfill volume required for a community's solid waste — Wastes with Fuel Value

A community of 110,544 people generates 3.0 kg/person/day of MSW and diverts 30% through recycling. Compacted in place at 714 kg/m³ with 20% additional daily cover volume, find the airspace needed for 22 years.

Given

Population = 110,544
Generation = 3.0 kg/cap/d
Diversion = 0.30
Compacted density = 714 kg/m³
Cover = 20%, design life 22 yr

Find

Annual and total landfill airspace

Start with the thinking

Only the landfilled fraction consumes airspace — diverted material is subtracted first.
Daily cover soil is real volume and must be added to the waste volume.

Step-by-step solution

Formula
$l an df i l l e d ma ss = p o p u l a t i o n \times r a t e \times (1 - d i v er s i o n)$
Substituting
$110544 (3.0) (1 - 0.30) = 232, 142 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 232, 142/714 = 325.1 m^{3} / d a y$
Annual with cover
$325.1 \times 365 \times 1.20 = 142, 407 m^{3} / y r$
Design life
$142, 407 \times 22 = 3, 132, 947 m^{3}$

Answer: ≈ 142,407 m³/yr, or 3,132,947 m³ over 22 years

Why the other options are there

3,729,699 m³ (diversion and cover ignored)
1,864,103,472 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 4

BOD exerted after t days — Wastes with Fuel Value (2)

A wastewater has ultimate BOD L₀ = 283 mg/L and k = 0.27 /day (base e). How much BOD is exerted in 9 days?

Given

L₀ = 283 mg/L
k = 0.27 /day
t = 9 days

Find

BOD_t

Start with the thinking

BOD exertion is first-order and approaches L₀ asymptotically.
Check whether k is base e or base 10.

Step-by-step solution

First-order
$B O D_{t} = L_{0} (1 - e^{(} - k t))$
Exponent
$k t = 0.27 (9) = 2.43$
Exponential
$e^{(} - 2.43) = 0.0880$
Substituting
$B O D_{t} = 283 (1 - 0.0880) = 258.1 m g / L$
Remaining
$L_{0} - B O D_{t} = 24.9 m g / L$

Answer: BOD_9 ≈ 258.1 mg/L

Why the other options are there

25 mg/L (remaining reported as exerted)
687.7 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 5

Activated sludge F/M ratio, aeration time and sludge production — Wastes with Fuel Value (2)

An activated sludge plant treats 3,351 m³/d with an influent BOD of 183 mg/L, effluent BOD 8 mg/L, MLSS 3,133 mg/L, and an aeration basin volume of 4,234 m³. With Y = 0.45 mg VSS/mg BOD, k_d = 0.06 d⁻¹ and an SRT of 6 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 3,351 m³/d
S₀ = 183 mg/L, S = 8 mg/L
X = 3,133 mg/L, V = 4,234 m³
Y = 0.45, k_d = 0.06 d⁻¹, SRT = 6 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 3351 (183) / (4234 \times 3133) = 0.046 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 4234/3351 = 1.2635 d = 30.32 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.45 (3351) (183 - 8) / (1 + 0.06 \times 6) m g / d$
Evaluate
$P_{x} = 194.0 k g V S S / d a y$

Answer: F/M = 0.05 d⁻¹, τ = 30.3 h, sludge = 194.0 kg/d

Why the other options are there

F/M = 195.7 (basin volume omitted)
P_x = 263.9 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 6

Landfill volume required for a community's solid waste — Wastes with Fuel Value (2)

A community of 104,755 people generates 2.5 kg/person/day of MSW and diverts 20% through recycling. Compacted in place at 606 kg/m³ with 15% additional daily cover volume, find the airspace needed for 18 years.

Given

Population = 104,755
Generation = 2.5 kg/cap/d
Diversion = 0.20
Compacted density = 606 kg/m³
Cover = 15%, design life 18 yr

Find

Annual and total landfill airspace

Start with the thinking

Only the landfilled fraction consumes airspace — diverted material is subtracted first.
Daily cover soil is real volume and must be added to the waste volume.

Step-by-step solution

Formula
$l an df i l l e d ma ss = p o p u l a t i o n \times r a t e \times (1 - d i v er s i o n)$
Substituting
$104755 (2.5) (1 - 0.20) = 209, 510 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 209, 510/606 = 345.7 m^{3} / d a y$
Annual with cover
$345.7 \times 365 \times 1.15 = 145, 119 m^{3} / y r$
Design life
$145, 119 \times 18 = 2, 612, 133 m^{3}$

Answer: ≈ 145,119 m³/yr, or 2,612,133 m³ over 18 years

Why the other options are there

2,839,275 m³ (diversion and cover ignored)
1,376,480,700 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 7

BOD exerted after t days — Wastes with Fuel Value (3)

A wastewater has ultimate BOD L₀ = 178 mg/L and k = 0.24 /day (base e). How much BOD is exerted in 3 days?

Given

L₀ = 178 mg/L
k = 0.24 /day
t = 3 days

Find

BOD_t

Start with the thinking

BOD exertion is first-order and approaches L₀ asymptotically.
Check whether k is base e or base 10.

Step-by-step solution

First-order
$B O D_{t} = L_{0} (1 - e^{(} - k t))$
Exponent
$k t = 0.24 (3) = 0.72$
Exponential
$e^{(} - 0.72) = 0.4868$
Substituting
$B O D_{t} = 178 (1 - 0.4868) = 91.4 m g / L$
Remaining
$L_{0} - B O D_{t} = 86.6 m g / L$

Answer: BOD_3 ≈ 91 mg/L

Why the other options are there

87 mg/L (remaining reported as exerted)
128.2 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 8

Activated sludge F/M ratio, aeration time and sludge production — Wastes with Fuel Value (3)

An activated sludge plant treats 12,271 m³/d with an influent BOD of 167 mg/L, effluent BOD 16 mg/L, MLSS 3,122 mg/L, and an aeration basin volume of 2,982 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.05 d⁻¹ and an SRT of 16 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 12,271 m³/d
S₀ = 167 mg/L, S = 16 mg/L
X = 3,122 mg/L, V = 2,982 m³
Y = 0.65, k_d = 0.05 d⁻¹, SRT = 16 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 12271 (167) / (2982 \times 3122) = 0.220 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 2982/12271 = 0.2430 d = 5.83 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.65 (12271) (167 - 16) / (1 + 0.05 \times 16) m g / d$
Evaluate
$P_{x} = 669.1 k g V S S / d a y$

Answer: F/M = 0.22 d⁻¹, τ = 5.8 h, sludge = 669.1 kg/d

Why the other options are there

F/M = 656.4 (basin volume omitted)
P_x = 1,204 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 9

Landfill volume required for a community's solid waste — Wastes with Fuel Value (3)

A community of 133,163 people generates 2.2 kg/person/day of MSW and diverts 35% through recycling. Compacted in place at 567 kg/m³ with 20% additional daily cover volume, find the airspace needed for 27 years.

Given

Population = 133,163
Generation = 2.2 kg/cap/d
Diversion = 0.35
Compacted density = 567 kg/m³
Cover = 20%, design life 27 yr

Find

Annual and total landfill airspace

Start with the thinking

Only the landfilled fraction consumes airspace — diverted material is subtracted first.
Daily cover soil is real volume and must be added to the waste volume.

Step-by-step solution

Formula
$l an df i l l e d ma ss = p o p u l a t i o n \times r a t e \times (1 - d i v er s i o n)$
Substituting
$133163 (2.2) (1 - 0.35) = 190, 423 k g / d a y$
Formula
$V = ma ss / d e n s i t y$
Substituting
$V = 190, 423/567 = 335.8 m^{3} / d a y$
Annual with cover
$335.8 \times 365 \times 1.20 = 147, 099 m^{3} / y r$
Design life
$147, 099 \times 27 = 3, 971, 682 m^{3}$

Answer: ≈ 147,099 m³/yr, or 3,971,682 m³ over 27 years

Why the other options are there

5,091,899 m³ (diversion and cover ignored)
1,876,619,552 m³ (mass reported as volume)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Example 10

BOD exerted after t days — Wastes with Fuel Value (4)

A wastewater has ultimate BOD L₀ = 255 mg/L and k = 0.16 /day (base e). How much BOD is exerted in 3 days?

Given

L₀ = 255 mg/L
k = 0.16 /day
t = 3 days

Find

BOD_t

Start with the thinking

BOD exertion is first-order and approaches L₀ asymptotically.
Check whether k is base e or base 10.

Step-by-step solution

First-order
$B O D_{t} = L_{0} (1 - e^{(} - k t))$
Exponent
$k t = 0.16 (3) = 0.48$
Exponential
$e^{(} - 0.48) = 0.6188$
Substituting
$B O D_{t} = 255 (1 - 0.6188) = 97.2 m g / L$
Remaining
$L_{0} - B O D_{t} = 157.8 m g / L$

Answer: BOD_3 ≈ 97 mg/L

Why the other options are there

157.8 mg/L (remaining reported as exerted)
122.4 mg/L (linear decay assumed)

Reference: FE Reference Handbook — Environmental Engineering → Wastes with Fuel Value

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Wastes with Fuel Value contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.